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I am given the set volume for a cylindrical container and separating pricing for the material used to construct the container, the base and the side wall. The ends of the container cost $0.05 per cm2 and the side walls $0.04 cm2, and the volume is 400 mL. The question asks: What is the best design for the container i.e., measurements to ensure that cost is at a minimum.

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my working:

V = 400 mL

Ends costs $0.05 per cm2

Side wall costs $0.04 cm2

Area of ends = [tex]2 \pi r^2[/tex]

Area of side wall [tex]2 \pi r h[/tex]

V = 400, therefore since V = [tex]\pi r^2 h[/tex], [tex]h = \frac{400}{\pi r^2}[/tex] (1)

Cost for one container = [tex]0.05 (2 \pi r^2) + 0.04 (2 \pi r h)[/tex] (2)

Therefore substituting (1) into (2), cost = [tex]0.05 (2 \pi r^2) + 0.04 (2 \pi r \frac{400}{\pi r^2})[/tex]

Differentiating the cost function gives: [tex]\frac {dC}{dr} = 0.2 \pi r + \frac {96}{r^2}[/tex]

The value of [tex]r[/tex] where there is no rate of change = -5.34601847

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I am not sure if what I have done after and including the differentiation is correct. Since I have got a negative answer for the value of r, I am thinking that what I have done is wrong, as you cannot have a negative radius. Would the second derivative help with the sloving this equation, if so how?

Many thanks,

Pavadrin

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# Homework Help: Optimisation of a cylindrical container

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